Solutions to Exam 1, Math 10560
1. The function f (x) = e2x + x3 + x is one-to-one (there is no need to check
this). What is (f −1 )0 (2 + e2 )?
Solution. Because f (x) is one-to-one, we know the inverse function exists.
1
Recall that (f −1 )0 (a) = f 0 (f −1
. So our first step is to find f −1 (2 + e2 ). By
(a))
definition of the inverse functions f −1 (2+e2 ) = x if and only if f (x) = 2+e2 .
So we solve for x:
f (x) = 2 + e2
e2x + x3 + x = 2 + e2 .
Comparing coefficients, we get x = 1.
The next step is to calculate f 0 (x):
f (x) = e2x + x3 + x
⇒ f 0 (x) = 2e2x + 3x2 + 1.
So f 0 (f −1 (2 + e2 )) = f 0 (1) = 2e2 + 3 + 1 = 2e2 + 4. Putting this into the
1
, we get:
equation (f −1 )0 (a) = f 0 (f −1
(a))
(f −1 )0 (2 + e2 ) =
1
.
+4
2e2
2. Use logarithmic differentiation to compute the derivative of the function
y=
2x (x3 + 2)
√
.
x−1
Solution. Taking the natural log of both sides of the equation, we get:
x 3
2 (x + 2)
√
ln y = ln
x−1
= ln(2x ) + ln(x3 + 2) − ln((x − 1)1/2 )
1
= x ln 2 + ln(x3 + 2) − ln(x − 1).
2
1
2
Differentiating both sides we obtain:
1 dy
3x2
1
·
= ln 2 + 3
−
y dx
x + 2 2(x − 1)
!
dy
3x2
1
⇒
.
= y ln 2 + 3
−
dx
x + 2 2(x − 1)
Thus,
!
3x2
1
ln 2 + 3
.
−
x + 2 2(x − 1)
2x (x3 + 2)
dy
= √
dx
x−1
3. Compute the integral
Z
1
5
1
(1 +
dx.
x2 ) tan−1 (x)
Solution. We use u-substitution with u = tan−1 (x) ⇒ du =
must also change the bounds of integration:
dx
.
1+x2
We
x = 5 → u = tan−1 (5),
x = 1 → u = tan−1 (1) = π/4.
Doing these substitutions, the integral changes to:
Z
tan−1 (5)
π/4
tan−1 (5)
1
du = ln |u|
= ln | tan−1 (5)| − ln(π/4).
u
π/4
Note that tan−1 (5) will be positive because 5 > 0, so the absolute value sign
isn’t needed. So,
Z 5
1
dx = ln(tan−1 (5)) − ln(π/4).
−1
2
1 (1 + x ) tan (x)
3
4. Find the derivative of (x2 + 1)x
2 +1
.
Solution. Since there is a variable in both the base and the exponent, we
2
need logarithmic differentiation. Let y = (x2 + 1)x +1 . Then
2
ln y = ln (x2 + 1)x +1
= (x2 + 1) · ln (x2 + 1).
Differentiating,
1 0
2x
y = 2x ln (x2 + 1) + (x2 + 1) 2
y
x +1
= 2x ln (x2 + 1) + 2x
= 2x ln (x2 + 1) + 1 .
Multiplying both sides by y = (x2 + 1)x
y 0 = (x2 + 1)x
2 +1
2 +1
, we obtain
2x ln (x2 + 1) + 1 .
5. Which of the following is true about y = f (x) = x ln(x), x > 0?
(a) The function is decreasing for 0 < x < 1e , increasing for x > 1e , and
concave up for all x > 0.
(b) The function is increasing for all x and concave up for all x > 0.
(c) The function is decreasing for 0 < x < e, increasing for x > e, and
concave up for all x > 0.
(d) The function is decreasing for 0 < x < 1, increasing for x > 1, and
concave up for all x > 0.
(e) The function is concave down for all x > 0.
Solution. Consider the first derivative
y 0 = ln x + 1.
We note that
1
ln x + 1 < 0 ⇐⇒ ln x < −1 ⇐⇒ x < e−1 = .
e
So the function y is decreasing on the interval 0 < x < 1e . Similarly,
1
,
e
so the function y is increasing on the interval x > 1e . To determine concavity,
we look at the second derivative
1
y 00 = .
x
ln x + 1 > 0 ⇐⇒ ln x > −1 ⇐⇒ x > e−1 =
4
Since
1
x
is positive for x > 0, the function y is concave up for all x > 0.
Answer:
The function is decreasing for 0 < x < 1e ,
increasing for x > 1e , and concave up for all
x > 0.
6. Compute the following definite integral.
Z log5 (6)
5x
√
dx.
1 + 5x
0
Solution. We choose the substitution u = 5x + 1. Then du = 5x ln 5 dx and
Z log5 (6)
Z 7
5x
1
1
√
√ du
dx =
x
ln 5 2
u
1+5
0
√ 7
2 u
=
ln 5 2
√
√ 2 7− 2
=
.
ln 5
5
7. Find the limit
ln(1 + 2x)
.
sin(x)
lim
x→0
Solution. Plugging x = 0 into the expression ln(1+2x)
sin(x) we see that the limit
0
is of indeterminate form 0 . We apply L’Hospital’s Rule:
d
[ln(1 + 2x)]
ln(1 + 2x)
= lim dx
lim
d
x→0
x→0
sin(x)
[sin(x)]
dx
= lim
x→0
2
1+2x
cos(x)
2
x→0 (cos(x))(1 + 2x)
2
=
cos(0) · 1
= lim
= 2.
8. Find the integral
Z
1
ex x2 dx.
0
Solution. We need to use integration by parts, with u = x2 , dv = ex dx.
Then du = 2xdx and v = ex . (We make this choice because we want to
decrease the power of x.) Now,
Z 1
Z 1
1
x 2
2 x
e x dx = x e − 2
xex dx
0
0
0
Z 1
1
= (e − 0) − 2
xex dx
0
Z 1
=e−2
xex dx.
0
6
R1
To evaluate the integral 0 xex dx, we again need to use integration by parts.
This time we choose u = x, dv = ex dx. So du = dx and v = ex . Thus,
Z 1
1 Z 1
x
x
ex dx
xe dx = xe −
0
0
0
1
= (e − 0) − ex 0
= e − (e − 1)
= 1.
Substituting this back into the above we obtain
Z 1
Z 1
xex dx
ex x2 dx = e − 2
0
0
=e−2·1
= e − 2.
Remark: IfRyou choose to drop the bounds and first evaluate the indefinite integral ex x2 dx, this is fine as long as your solution is written in a
mathematically correct way (i.e. do not have a definite integral equal to
an indefinite integral). An example of how to do this would be as folllows.
Applying integration by parts twice (with the same choices for u and dv as
above) we obtain:
Z
Z
x 2
2 x
e x dx = x e − 2 xex dx
Z
2 x
x
x
= x e − 2 xe − e dx
= x2 ex − 2 (xex − ex ) + C
= x2 ex − 2xex + 2ex + C.
Therefore,
Z
0
1
1
ex x2 = x2 ex − 2xex + 2ex 0
1
1
= (e1 − 2e
+
2e
) − (0 − 0 + 2e0 )
= e − 2.
7
9. Compute the integral
Z
π/4
tan3 (x) sec3 (x) dx.
0
R
Solution. This is an integral of the form secm x tann x dx. Since both m
and n are odd, we use the u-substitution u = sec x, du = sec x tan x and use
the trigonometric identity 1 + tan2 x = sec2 x to rewrite tan2 x as sec2 x − 1.
Putting this all together, we obtain:
Z
π/4
tan3 x sec3 x dx =
π/4
Z
0
(tan2 x sec2 x)(sec x tan x) dx
0
π/4
Z
(sec2 x − 1)(sec2 x)(sec x tan x) dx
=
0
sec(π/4)
Z
(u2 − 1) · u2 du
=
sec(0)
√
2
Z
u4 − u2 du
=
1
√ 2
u5 u3
=
−
5
3 1
√ 5
√ ! ( 2)
( 2)3
1 1
=
−
−
−
5
3
5 3
25/2 − 1 23/2 − 1
−
.
5
3
=
10. Compute the integral
Z
0
2
x
√ dx.
x2 + 3 x
Hint: a rationalizing substitution might help.
Solution. We can rewrite our integral as
Z 2
Z 2
x
x
√
dx
=
dx
2+ 3x
1/3 (x5/3 + 1)
x
x
0
0
Z 2
x2/3
=
dx.
5/3 + 1
0 x
8
We now see that we can use the u-substitution u = x5/3 + 1, du = 53 x2/3 dx.
Next, we change the bounds of integration:
x = 0 ⇒ u = 05/3 + 1 = 1,
√
3
x = 2 ⇒ u = 25/3 + 1 = 32 + 1.
Putting this together, we obtain
Z
0
2
x
√ =
2
x + 3x
Z
0
2
x2/3
dx
x5/3 + 1
3
Z √
32+1
du
3
=
5 1
u
√
3
32+1
3
= ln |u|
5
1
√
3 ln( 3 32 + 1)
=
.
5
11. The population (in thousands (= th.)) of UNCland had an initial
value of 100 th. at time t = 0 days. The population of UNCland has been
decreasing since then at a rate proportional to its size. The population of
UNCland was 60 th. at time t = 20 days. Let P (t) denote the size of the
population of UNCland t days after t = 0.
(a)
Find an expression for P (t) in the form
P (t) = Cekt ,
where C and k are constants. (You should NOT attempt to approximate
numbers such as ln(2) etc.. in decimal or fractional form. )
Solution. At t = 0, we have: P (0) = 100 = Cek·0 = Ce0 = C, so C = 100.
At t = 20, we have:
P (20) = Ce20k
60 = 100e20k
60
3
= = e20k ,
100
5
9
Taking the natural logarithm of both sides:
3
= ln(e20k )
5
ln(3) − ln(5) = 20k
ln
ln(3) − ln(5)
= k.
20
Answer:
C = 100th., k =
ln 3 − ln 5
20
(b) When will the population of UNCland equal 10 th. ?
(You may write your answer in terms of logarithms. )
Solution. We want to know for which t we get P (t) = 10:
P (t) = 100ekt = 10
⇒e
ln 3−ln 5
·t
20
=
10
1
=
100
10
Take the natural logarithm of both sides:
ln 3 − ln 5
· t = ln(1/10) = − ln 10
20
−20 ln(10)
20 ln(10)
t=
=
ln(3) − ln(5)
ln(5) − ln(3)
So the population of UNCland is 10 th. when
t=
Equivalent Answers:
20 ln(10)
days.
ln(5) − ln(3)
20 ln(1/10)
ln(3/5)
days,
20 ln(10)
ln(5/3)
days.
12. Compute the integral
Z
x2 + 1
dx.
x3 + x2
Solution. We proceed using partial fraction decomposition. First, we factor
the denominator
x3 + x2 = x2 (x + 1).
10
Since the linear factor x occurs twice, the partial fraction decomposition is
x2 + 1
C
A
B
= + 2+
.
2
x (x + 1)
x
x
x+1
Multiplying both sides by x2 (x + 1), we get
x2 + 1 = Ax(x + 1) + B(x + 1) + Cx2
= (A + C)x2 + (A + B)x + B.
Equating coefficients
1=B
0=A+B
1=A+C
we obtain A = −1, B = 1, and C = 2. Now, returning to the integral,
Z
Z
1
1
2
x2 + 1
dx = − + 2 +
dx
x2 (x + 1)
x x
x+1
= − ln |x| −
1
+ 2 ln |x + 1| + C.
x
13. Compute
Z
√
1
dx
x2 − 2x + 2
Present your answer as a function of the variable x. Note the formula sheet
at the back of the exam may be helpful in working out your final answer.
Solution. Completing the square we see that x2 − 2x + 2 = x2 − 2x + 12 −
12 + 2 = (x − 1)2 + 1. Thus we can rewrite our integral as
Z
1
√
dx =
2
x − 2x + 2
Z
1
p
(x − 1)2 + 1
Z
1
√
=
du
2
u +1
dx
(Substitution: u = x − 1, du = dx)
Next, we want to make a trigonometric substitution. Because our expression
is of the form u2 + a2 , we use the substitution u = tan θ, − π2 ≤ θ ≤ π2 . So
du = sec2 θ dθ, and our integral becomes
11
Z
1
√
dx =
x2 − 2x + 2
Z
Z
=
√
√
1
u2 + 1
1
tan2 θ
du
+1
sec2 θ
√
dθ
=
2θ
sec
Z
= sec θdθ
· sec2 θ dθ
Z
= ln | sec θ + tan θ| + C.
Next we need to substitute back in for θ. We had u = tan θ, where − π2 ≤
θ ≤ π2 . Further, since − π2 ≤ θ ≤ π2 , we have sec θ = cos1 θ > 0. Now, sec2 θ =
√
1 + tan2 θ = 1 + u2 , so taking the positive square root, sec θ = u2 + 1.
Now,
Z
1
√
= ln | sec θ + tan θ| + C
2
x − 2x + 2
p
= ln | u2 + 1 + u| + C
Finally, we had u = x − 1 and
back in we obtain
Z
√
√
u2 + 1 =
√
x2 − 2x + 2, so substituting this
p
1
dx = ln | x2 − 2x + 2 + x − 1| + C.
x2 − 2x + 2